7/8b-35=3/4b+32

Solution for 7/8b-35=3/4b+32 equation:

7/8b-35=3/4b+32

We move all terms to the left:

7/8b-35-(3/4b+32)=0


Domain of the equation: 8b!=0

b!=0/8

b!=0

b∈R



Domain of the equation: 4b+32)!=0

b∈R


We get rid of parentheses

7/8b-3/4b-32-35=0

We calculate fractions


28b/32b^2+(-24b)/32b^2-32-35=0

We add all the numbers together, and all the variables

28b/32b^2+(-24b)/32b^2-67=0

We multiply all the terms by the denominator


28b+(-24b)-67*32b^2=0

Wy multiply elements

-2144b^2+28b+(-24b)=0

We get rid of parentheses

-2144b^2+28b-24b=0

We add all the numbers together, and all the variables

-2144b^2+4b=0

a = -2144; b = 4; c = 0;
Δ = b2-4ac
Δ = 42-4·(-2144)·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4}{2*-2144}=\frac{-8}{-4288}
=1/536
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4}{2*-2144}=\frac{0}{-4288}
=0
$